0
$\begingroup$

Please see this proposition.

Let $G$ be a group with subgroups $G_1,..., G_n$ such that $\mathcal{D}_1,\mathcal{D}_2,\mathcal{D}_3$ hold, then the mapping \begin{align*} \alpha: {\huge\times}_{i=1}^n G_i\rightarrow G\text{ with }(g_1,...,g_n)\mapsto g_1\cdots g_n \end{align*} is a isomorphism.

$\mathcal{D}_1~~~~G=G_1\cdots G_n;$

$\mathcal{D}_2~~~~G_i\trianglelefteq G;$

$\mathcal{D}_3~~~~G_i\cap \prod_{j\neq i}G_j=1.$

It's a rather common proposition. But I'm stuck on one point when I tried to go through the proof.

Question:

$G_i\cap \prod_{j\neq i}G_j$ yields $[G_i,G_j]\leq G_i\cap\prod_{j\neq i}G_j=1$, which means for all $1\leq i,j\leq n$, $G_i,G_j$ centralizes each other. Since all $g\in G$ can be written as $g=g_1\cdots g_n$, then we'll have $$\boldsymbol{ gg^*=(g_1\cdots g_n)(g^*_1\cdots g^*_n)=(g^*_1\cdots g^*_n)(g_1\cdots g_n)=g^*g,}$$ where $g,g^*$ are arbitrary elements in $G$.

But it is utterly ridiculous, isn't it?

I'm a beginner in group theory, would you please give me some help and point out where I misunderstood? Thanks!

$\endgroup$
  • $\begingroup$ You seem to be assuming that $g_i$ commutes with $g_i^*$. That's not necessarily true if $G_i$ is nonabelian. $\endgroup$ – Bungo Mar 13 '18 at 5:38
  • $\begingroup$ @Bungo Ah, yes! I feel like a fool... Thanks a lot. $\endgroup$ – Math Mar 13 '18 at 5:40
  • $\begingroup$ Should I delete this post? It this question valuable? $\endgroup$ – Math Mar 13 '18 at 5:41
  • $\begingroup$ No problem. The rest of your argument, namely that elements of $G_i$ commute with elements of $G_j$ whenever $i \neq j$, is correct. $\endgroup$ – Bungo Mar 13 '18 at 5:41
  • $\begingroup$ Someone might find it useful. I'll go ahead and write a quick answer so it doesn't have to stay open. $\endgroup$ – Bungo Mar 13 '18 at 5:43
1
$\begingroup$

Your argument is almost correct: elements of $G_i$ commute with elements of $G_j$ provided that $i \neq j$. However, elements of $G_i$ need not commute with each other, since $G_i$ is not assumed abelian.

$\endgroup$
  • $\begingroup$ So the positions of $G_i$ can change, am I right? $\endgroup$ – Math Mar 13 '18 at 5:56
  • 1
    $\begingroup$ If you mean that we can reorder the $G_i$'s in the product $G = G_1 G_2 \cdots G_n$, that is correct. In fact, you only need hypothesis $\mathcal D_2$ in order to do this: normal subgroups $H$ and $K$ always satisfy $HK = KH$, and by induction this holds for any finite number of normal subgroups. Your result is stronger: not only do the subgroups $G_i$ commute as sets, their individual elements also commute. $\endgroup$ – Bungo Mar 13 '18 at 5:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.